Problem: Let $h(x)=\ln(x)\cos(x)$. $h'(x)=$
Answer: $h(x)$ is the product of two, more basic, expressions: $\ln(x)$ and $\cos(x)$. Therefore, the derivative of $h$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}(\ln(x)\cos(x)) \\\\ &=\dfrac{d}{dx}(\ln(x))\cos(x)+\ln(x)\dfrac{d}{dx}(\cos(x))&&\gray{\text{The product rule}} \\\\ &=\dfrac{1}{x}\cdot \cos(x)+\ln(x)\cdot (-\sin(x))&&\gray{\text{Differentiate }\ln(x)\text{ and }\cos(x)} \\\\ &=\dfrac{\cos(x)}{x}-\ln(x)\sin(x)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $h'(x)=\dfrac{\cos(x)}{x}-\ln(x)\sin(x)$ or any other equivalent form.